Ile powstanie kwasu siarkowego (IV) jeżeli w wodzie rozpuszczono 256g tlenku siarki (IV)?
mSO2=64u
mH2SO3=82u
SO2 + H2O---->H2SO3
64g SO2------82g H2SO3
256g SO2-------xg H2SO3
x = 328g H2SO3
SO2+H2O----->H2SO3
64g SO2 --------- 82g H2SO3
256g SO2 -------- x
x=328g H2SO3
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mSO2=64u
mH2SO3=82u
SO2 + H2O---->H2SO3
64g SO2------82g H2SO3
256g SO2-------xg H2SO3
x = 328g H2SO3
SO2+H2O----->H2SO3
64g SO2 --------- 82g H2SO3
256g SO2 -------- x
x=328g H2SO3